如何从Linear SVM手动使用拟合模型进行预测？

Question

The scikit.learn function .predict from the library LinearSVC performs the prediction using test samples.

LinearSVM_cl.fit(X_train , Y_train) 

And the prediction with

Y_pred_LinearSVM = LinearSVM_cl.predict(X_test)  

However, I need to know which parameters from the fit model is used to predict a test samples, .coef_? .intercept_?

The dataset for the model is 20000 rows and 8 columns obtaining with 8 classes:

.coef ->

 array([[-1.20185887, -0.62510767, -0.92739275, -0.08900084, -1.11164502,
    -0.56442702,  1.92045989, -0.56706939],
   [ 0.75386897,  0.9672828 , -2.10451063,  0.53552943, -0.10476675,
     0.32058617, -0.30133408, -1.01478727],
   [ 0.35032536, -0.38405342,  0.25462054,  0.47577302, -0.55000734,
     0.01134098, -0.14534849,  1.14597475],
   [-0.08888566, -0.08272116,  0.84141105,  0.22040919,  0.27763948,
     0.57907834, -0.70631803, -0.1017982 ],
   [ 0.14319018,  0.03329494,  1.52575489,  0.58355648,  1.24454465,
    -0.92758526,  1.01315744, -0.51935599],
   [-0.33712774, -0.7826993 , -1.00810522,  0.20346304,  3.67215014,
     0.93187058, -0.26441527, -0.5351838 ],
   [-0.70416157, -2.38388785, -1.24720653,  0.43291862,  3.91473792,
     2.7596399 , -0.63503461, -0.43277051],
   [-0.14921538, -0.03871313, -0.19896247,  0.08522851,  0.29347373,
     0.1332059 , -0.10875692, -0.01503476]])

.intercept ->

array([-0.43454897,  0.05659295, -0.95980815, -1.36353241, -3.05042133,
   -2.93684622, -3.35757856, -1.14034588])

And example of test sample is

   0.7622999 0.514543 0.2195486 0.453202 0.2585706 0.6295224 0.4999675 0.1960128

How can I predict the test sample manually (without using the built .predict function from the library).

Answer 1

Note your coef as $W$ and your intercept as $b$ and your new data point as $x$. Your class prediction is simply:

$c = \\arg \\max_i{W_i \\cdot x + b} $

So you just apply matrix multiplication, add the bias vector and pick the index of the maximal entry.